If $a>2b>0$ then the positive value of m for which $y=mx-b\sqrt{1+m^2}$ is a common tangent to

$x^2+y^2=b^2$ and $(x-a{)}^2+y^2=b^2$ is

Clearly, $y=mx-b\sqrt{1+m^2}$ is a tangent to

$x^2+y^2=b^2$ It will touch the circle $(x-a{)}^2+y^2=b^2$ if

$\begin{array}{l}\left|\frac{\left|ma-0-b\sqrt{1+m^2}\right|}{\sqrt{m^2+1}}\right|=b\\ \Rightarrow \left|ma-b\sqrt{1+m^2}\right|=b\sqrt{1+m^2}\\ \Rightarrow ma-b\sqrt{1+m^2}=\pm b\sqrt{1+m^2}\\ \Rightarrow ma=2b\sqrt{1+m^2}\text{ or, }ma=0\\ \Rightarrow m^2{a}^2=4b^2+4b^2{m}^2\end{array}$

$\Rightarrow m=\pm \frac{2b}{\sqrt{a^2-4b^2}}\Rightarrow m=\frac{2b}{\sqrt{a^2-4b^2}}[\because m>0]$